Calculation for Multiple Effect Evaporator In doing calculation for a multiple effect evaporator system, the value to be obtained are usually the area of heating surface in each effect, the kg of steam per hour to be supplied, and the amount of vapor leaving each effect, especially the last effect. The given or known value are usually as follows : (1) steam pressure to first effect, (2) final pressure in vapor space of the last effect, (3) feed conditions and flow to first effect, (4) the final concentration in the liquid leaving the last effect, (5) physical properties such as enthalpies and/or heat capacities of the liquid and vapors, and (6) the overall heat-transfer coefficient in each effects. The calculation are done using mass balances, energy balances, and the heat transfer rate equations q = U A ∆T for each effect. Step-by-Step Calculation Methods for Triple Effect Evaporators 1. From the known outlet concentration and pressure in the last effect, determined the BP in the last effect.
2. Determine the total amount of vapor evaporated by an overall mass balance. For this first trial apportion this total amount of vapor among the three effects. (Usually V1 = V2 = V3 is assumed for the first trial). Make a total mass balance on effects 1, 2, 3 to obtain L1, L2, and L3. Then calculate the solids concentration in each effect by a solid balance on each effect. 3. Using equation 1/U1 ∆T1 = Σ ∆T ──────────── 1/U1 + 1/U2 + 1/U3 Estimate the temperature drops ∆T1, ∆T2, and ∆T3 in the three effects. Any effect that has an extra heating load, such as cold feed, requires a proportionately larger ∆T. Then calculate the BP in each effect. ( If BPR is present, estimated the pressure in the effect 1 and 2 and determinated the BPR in each of the three effect. Only a crude pressure estimated is needed since BPR is almost independent of pressure. Then the Σ ∆T available for heat transfer without the superheated is obtained by subtracting the sum of all three BPRs from overall ∆T of TS – T3. Using equation 1/U1 ∆T1 = Σ ∆T ────────────
1/U1 + 1/U2 + 1/U3 estimate ∆T1, ∆T2, and ∆T3. Then calculate the BP in each effect.) 4. Using energy and mass balances in each effect, calculate the amount vaporized and the flows of liquid in each effect.
If the amounts vaporized differ
aprreciably from those assumed in step 2, then step 2, 3, 4 can be repeated using the amounts of evaporation just calculated.
(In step 2 only solid balance is
repeated.) 5. Calculate the value of q transffered in each effect. Using the rate equation q = U A ∆T for each effect, calculate area A1, A2, A3.
Then calculate the average
value Am by Am = (A1 + A2 + A3)/3 If these A are reasonably close to each other, the calculation are complete and second trial is not needed. If these area are not nearly equal, a second trial should be performed on each effect. 6. To start trial 2, use new values of L 1, L2, L3, V1, V2, and V3
calculate the energy balances in step 4 and
calculate the new solids concentrations in each effect by a solids balances on each effect. 7. Obtained new values of ∆T1’, ∆T2’, and ∆T3’ from ∆T1 A1 ∆T1 A1
∆T1’ = ──── Am
∆T2’ = ──── Am
∆T1 A1 and ∆T3 = ───── Am ’
Then sum ∆T1’, ∆T2’, and ∆T3’ must equal the original Σ ∆T. If not, proportionately readjust all ∆T ’ value so that this is so. Then calculate the BP in each effect. ( If BPR is present, then using the new concentrations from step 6, determine the new BPRs in the three effects.
This gives new value of Σ ∆T available for
heat transfer by subtracting the sum of all of three BPRs from overall ∆T. Calculate the new of ∆T’ by equation ∆Ti Ai ∆Ti = ──── Am ’
to obtained a better estimeted of the ∆T’ 8. Using the new ∆T’ values from step 7, repeat the calculations starting with step 4. Two trial are usually sufficient so that the areas are reasonably close to being equal. Example
Evaporation of Sugar Solution in a Triple-Effect Evaporator A triple effect forward feed evaporator is being used to evaporate a sugar solution containing 10 mass % solids to a concentrated solution 50 mass %.
The BPR os
solutions (independent pressure) can be estimated from BPR 0C = 1.78 x + 6.22x2 where x is mass fraction of sugar in solution.
Saturated steam at 205.5 kPa (121.1 0C
saturation temperature) is being used.
The pressure in
the vapor space of the third effect is 13.4 kPa. The feed rate is 22680 kg/h at 26.7 0C. The heat capacity of the liquid solutions is Cp = 4.19 – 2.35 x kJ/kg.0K. The heat solution is considered to be negligible. The coefficients of the heat transfer have been estimated as U 1 = 3123, U2 = 1987, and U3 = 1136 W/m2.0K. If each effect has the same surface area, calculate the area, the steam rate used, and the steam economy. Solution The process flow diagram is given in fig. Following the eight step outlined, the calculation as follows. Step 1. For the 13.4 kPa, the saturation temperature is 51.67 0C from the steam tables. Using the equation for BPR for evaporator no. 3 with x = 0,5, BPR3 = 1.78x + 6.22x2 = 1.78 (0.5) + 6.22 (0.5)2
= 2.450C T3 = 51.67 + 2.45 = 54.120C Step 2 Making an overall and a solids balance to calculate the total amount vaporized (V1 + V2 + V3) and L3 F = 22.680 = L3 + (V1 + V2 + V3) F xF = 22.680 (0.1) = L3 (0.5) L3 = 4536 kg/h (V1 + V2 + V3) = 22.680 - 4536 = 18.144 kg/h Assuming equal amount vaporized in each effect, V1 = V2 = V3 = 6048 kg/h Making total mass balance on effect 1, 2, 3 and solving (1) F = 22.680 = V1 + L1, L1 = 16.632 kg/h (2) L1 = 16.632 = V2 + L2, L2 = 10.584 kg/h (3) L2 = 10.584 = V3 + L3, L3 = 4536 kg/h Making a solids balance on effect 1, 2, and 3 and solving for x (1)
F xF = L1 x1, 22.680 (0.1) = 16.632 (x1) x1 = 0.136
(2)
L1 x1 = L2 x2, 16.632 (0.136) = 10.584 (x2) x2 = 0.214
(3)
L2 x2 = L3 x3, 10.584 (0.214) = 4536 (x3) x2 = 0.50 (check balance)
Step 3 The BPR in each effect is calculated as follows:
(1)
BPR1 = 1.78x + 6.22x2 = 1.78 (0.136) + 6.22 (0.136)2 =
(2)
0.360C BPR2 = 1.78x + 6.22x2 = 1.78 (0.214) + 6.22 (0.214)2 =
(3)
0.650C BPR3 = 1.78x + 6.22x2 = 1.78 (0.5) + 6.22 (0.5)2 = 2.450C
Σ ∆T available = TS1 – T3 (saturation) – (BPR1 + BPR2 + BPR3) = 121.1 – 51.67 – (0.36 + 0.65 + 2.45) = 65.970C Using equation 1/U1 ∆T1 = Σ ∆T ──────────── 1/U1 + 1/U2 + 1/U3 65.97 (1/3132) = ──────────────── = 12.400C 1/3132 + 1/1987 + 1/1136 1/U2 ∆T2 = Σ ∆T ──────────── 1/U1 + 1/U2 + 1/U3 65.97 (1/1987) = ──────────────── = 19.500C 1/3132 + 1/1987 + 1/1136 1/U3 ∆T3 = Σ ∆T ──────────── 1/U1 + 1/U2 + 1/U3 65.97 (1/1136) = ──────────────── = 34.070C 1/3132 + 1/1987 + 1/1136
However, since a cold feed enters effect no.1, this effect requires more heat.
Increased ∆T1 and lowering ∆T2
and ∆T3 proportionately as a first estimate, ∆T1 = 15.560C ∆T2 = 18.340C ∆T3 = 32.070C To calculate the actual BP of the solution in each effect, (1) T1 = TS1 - ∆T1 = 121.1 – 15.56 = 105.540C TS1 = 121.10C (2) T2 = T1 – BPR1 - ∆T2 = 105.54 – 0.36 - 15.56 =
(3)
86.840C TS2 = T1 – BPR1 = 105.54 – 0.36 = 105.180C T3 = T2 – BPR2 - ∆T3 = 86.84 – 0.65 - 32.07 = 54.120C TS3 = T2 – BPR2 = 86.84 – 0.65 = 86.190C
The temperature in the three effect are as follows: Effect 1 Effect 2 Effect 3 TS1 = 121.10C T1 = 105.540C
TS2 = 105.18 T2 = 86.840C
TS3 = 86.19 T3 = 54.120C
TS4 = 51.67
Step 4 The heat capacity of the liquid in each effect is calculated from the equation Cp = 4.19 – 2.35 x. F : Cp = 4.19 – 2.35 (0.1) = 3.955 kJ/kg.0K L1 : Cp = 4.19 – 2.35 (0.136) = 3.869 kJ/kg.0K L2 : Cp = 4.19 – 2.35 (0.214) = 3.684 kJ/kg.0K L3 : Cp = 4.19 – 2.35 (0.50) = 3.015 kJ/kg.0K
The value of the enthalpy H of the various vapor streams relative to water at 00C as a datum are obtained from the steam tables as follows: Effect 1 : T1 = 105.540C, TS2 = 105.180C, BPR1 = 0.36, TS1 = 121.10C H1 = HS2 (saturation enthalpy) + 1.884 (0.36) (superheat) = 2684 + 1.884 (0.36) = 2685 kJ/kg λS1 = HS1 (vapor saturation enthalpy) - hS1 (liquid of condensation). Effect 2 T2 = 86.840C, TS3 = 105.180C, BPR2 = 0.65, H2 = HS3 (saturation enthalpy) + 1.884 (0.65) (superheat) = 2654 + 1.884 (0.36) = 2655 kJ/kg λS2 = H1 - hS2 = 2685 - 441 = 2244 kJ/kg (latent heat of condensation). Effect 3 T3 = 54.120C, TS4 = 51.670C, BPR3 = 2.45, H3 = HS4 + 1.884 (2.45) = 2595 + 1.884 (2.45) = 2600 kJ/kg λS3 = H2 - hS3 = 2655 - 361 = 2294 kJ/kg (latent heat of condensation). Flow relation to be used in energy balances are V1 = 22.680 – L1, V2 = L1 - L2, V3 = L2 - 4536 L3 = 4536 Write energy balance on each effect. Use 00C as a datum since the value of H of the vapors are realtive to 0 0C and note that (TF - 0)0C = (TF - 0)0K (1) F Cp (TF – 0) + S λS = L1 Cp (T1 – 0) + V1 H1
22.680 (3.955) (26.7 – 0) + S (2200) = L 1 (3.869) (2)
(105.54 – 0) + (22.680 – L1) (2685) L1 Cp (T1 – 0) + V1 λS2 = L2 Cp (T2 – 0) + V2 H2 L1 (3.869) (105.54 – 0) + (22.680 – L1) (2244) = L2
(3)
(3.684) (86.84 – 0) + (L1 – L2) (2655) L2 Cp (T2 – 0) + V2 λS3 = L3 Cp (T3 – 0) + V3 H3 L2 (3.684) (86.84 – 0) + (L1 – L2) (2294) = 4536 (3.015) (54.12 – 0) + (L2 – 4536) (2600)
Solving the last two equation simultaneously for L 1 and L2 and substituting into the first equation L1 = 17078 kg/h, L2 = 11.068, L3 = 4536 S = 8936, V1 = 5602, V2 = 6010, V3 = 6532 Step 5 Solving for the values of q in each effect and area q1 = S λS1 = (8936/3600) (2200 x 1000) = 5.460 106 W q2 = S λS2 = (5602/3600) (2244 x 1000) = 3.492 106 W q3 = S λS3 = (6010/3600) (2294 x 1000) = 3.830 106 W A1 = q1/U1 ∆T1 = 5.460 106/3123 (15.56) = 112.4 m2 A2 = q2/U2 ∆T2 = 3.492 106/1987 (18.34) = 95.8 m2 A3 = q3/U3 ∆T3 = 3.830 106/1136 (32.07) = 105.1 m2 The average area Am = 104.4 m2.
The areas differ from
from the average value less than 10% and a second trial is not really necessary. methods use. Step 6
However, a second trial will be
Making a new solids balance on effects 1, 2, and 3 using the new L1 = 17.078, L2 = 11.068, L3 = 4536 and solving for x, (1) 22.680 (0.1) = 17.078 (x1), x1 = 0.133 (2) 17.078 (0.133) = 11.068 (x2), x2 = 0.205 (3) 11.068 (0.205) = 4536 (x3), x3 = 0.50 (check balance) Step 7 The new BPR in each effect is then (1) (2) (3)
BPR1 = 1.78 x1 + 6.22 x12 = 1.78 (0.133) + 6.22 (0.133)2 = 0.350C BPR2 = 1.78 x2 + 6.22 x22 = 1.78 (0.205) + 6.22 (0.205)2 = 0.630C BPR3 = 1.78 x3 + 6.22 x32 = 1.78 (0.50) + 6.22 (0.50)2 = 2.450C
Σ ∆T available = 121.1 – 51.67 – (0.35 + 0.63 + 2.45) = 66.00C The new values for ∆T are obtained using equation ∆Ti Ai ∆Ti = ──── Am ’
∆T1 A1 15.56 (112.4) ∆T1 = ──── = ──────── = 16.770K = 16.770C Am 104.4 ∆T2 A2 18.34 (95.8) ’ ∆T2 = ──── = ──────── = 16.860K = 16.860C Am 104.4 ’
∆T3 A3 32.07 (105.1) ∆T3 = ──── = ──────── = 32.340K = 32.340C Am 104.4 ’
Σ ∆T = 16.77 + 16.86 + 32.34) = 65.970C These ∆T’ values are readjusted so that ∆T1’ = 16.77, ∆T2’ = 16.87, ∆T3’ = 32.36. Σ ∆T = 16.77 + 16.87 + 32.36 = 66.00C To calculate the actual boiling point of the solution in each effect (1) T1 = TS1 - ∆T1 = 121.1 – 16.67 = 104.330C (2) T2 = T1 – BPR1 - ∆T2 = 104.33 – 0.35 - 16.87 =
(3)
87.110C TS2 = T1 – BPR1 = 104.33 – 0.35 = 103.980C T3 = T2 – BPR2 - ∆T3 = 87.11 – 0.63 - 32.36 = 54.120C TS3 = T2 – BPR2 = 87.11 – 0.63 = 86.480C
Step 8 Following step 4, the heat capacity of the liquid is C p = 4.19 – 2.35 x F : Cp = 4.19 – 2.35 (0.1) = 3.955 kJ/kg.0K L1 : Cp = 4.19 – 2.35 (0.133) = 3.877 kJ/kg.0K L2 : Cp = 4.19 – 2.35 (0.205) = 3.708 kJ/kg.0K L3 : Cp = 4.19 – 2.35 (0.50) = 3.015 kJ/kg.0K The new enthalpy H are as follows in each effect. (1) H1 = HS2 + 1.884 (0C superheat) = 2682 + 1.884 (0.35) = 2683 kJ/kg λS1 = HS1 - hS1 = 2708 – 508 = 2200 kJ/kg
(2)
H2 = HS3 + 1.884 (0.63) = 2654 + 1.884 (0.63) =
(3)
2655 kJ/kg λS2 = H1 - hS2 = 2683 - 440 = 2243 kJ/kg H3 = HS4 + 1.884 (2.45) = 2595 + 1.884 (2.45) = 2600 kJ/kg λS3 = H2 - hS3 = 2655 - 361 = 2294 kJ/kg
Writing the energy balance on each effect, (1) F Cp (TF – 0) + S λS = L1 Cp (T1 – 0) + V1 H1 22.680 (3.955) (26.7 – 0) + S (2200) = L 1 (3.877) (2)
(104.33 – 0) + (22.680 – L1) (2683) L1 Cp (T1 – 0) + V1 λS2 = L2 Cp (T2 – 0) + V2 H2 L1 (3.877) (104.33 – 0) + (22.680 – L1) (2243) = L2
(3)
(3.708) (87.11 – 0) + (L1 – L2) (2655) L2 Cp (T2 – 0) + V2 λS3 = L3 Cp (T3 – 0) + V3 H3 L2 (3.708) (87.11 – 0) + (L1 – L2) (2293) = 4536
(3.015) (54.12 – 0) + (L2 – 4536) (2600) Solving, L1 = 17.005 kg/h, L2 = 10.952 kg/h, L3 = 4536 kg/h, S = 8960 kg/h,
V1 = 5675 kg/h,
V2 = 6053 kg/h,
V3 = 6416 kg/h. Following step 5, and solving for q in each effect and A q1 = S λS1 = (8960/3600) (2200 x 1000) = 5.476 106 W q2 = V1 λS2 = (5675/3600) (2243 x 1000) = 3.539 106 W q1 = V2 λS3 = (6053/3600) (2293 x 1000) = 3.855 106 W A1 = q1/U1 ∆T1 = 5.476 106/3123 (16.77) = 104.6 m2 A2 = q2/U2 ∆T2 = 3.539 106/1987 (16.87) = 105.6 m2 A3 = q3/U3 ∆T3 = 3.855 106/1136 (32.36) = 104.9 m2 The average area Am = 105.0 m2 to use in each effect. V1 + V2 + V3 Steam Economy = ──────────
S 5675 + 6053 + 6416 Steam Economy = ────────────── = 2.025 8960
